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When (2​S,3​R)-3-iodobutan-2-ol undergoes a substitution reaction with sodium azide the only organic product from the reaction is (2​S,3​R)-3-azidobutan-2-ol. Give a mechanism for the reaction.

My Attempt

Now I know this is not a normal $\mathrm{S_N1}$ or $\mathrm{S_N2}$ reaction since the stereochemistry remains the same and only one of the enantiomers are formed. I am guessing that at first there is some intramolecular reaction where the iodine atom attacks the other carbon bonded with the hydroxyl group. Then the azide substitutes the iodine atom.

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I can explain the reason why the configuration does not change by using neighboring group participation of the hydroxyl group. This is the mechanism that I propose:

  1. The hydroxyl group attacks C-3 and $\ce{I-}$ leaves.
  2. Then $\ce{N3-}$, the nucleophile, attacks C-3.
  3. Since two successive inversions occur at C-3, the net result is retention.

Mechanism

As you can see, the configuration has not changed.


@bon brought up a point: The formation of a mixture of products when an epoxide opening takes place. If the nucleophile were to attack C-2 instead,

Attack at alternative carbon

C-3 will still have the S configuration.


As noted by @Jan in the comments, the epoxide intermediate is symmetrical and both epoxide carbons are equivalent. This can be shown by a $C_2$ rotation:

Epoxide symmetry

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  • $\begingroup$ Epoxide formation under neutral conditions? Also, one would expect a mixture of products from the epoxide opening. $\endgroup$ – bon Feb 12 '16 at 16:55
  • $\begingroup$ @bon I am sorry if this is wrong. I read about retention of configuration because of neighboring group participation in jerry march's book. So I thought it would be the reason here. $\endgroup$ – Aditya Dev Feb 12 '16 at 16:56
  • $\begingroup$ I'm not saying it's definitely wrong but I'm just surprised (and intrigued). $\endgroup$ – bon Feb 12 '16 at 16:57
  • $\begingroup$ @bon: if the nucleophile were to attack at the third carbon, then the product will also have retention configuration. So Nu attack on 2nd and third carbon will produce the same compound. I just drew the possible products on my book. $\endgroup$ – Aditya Dev Feb 12 '16 at 16:59
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    $\begingroup$ The intermediate epoxide is $C_2$ symmetric. This renders the carbons homotopic and means that both attacks give the same product. $\endgroup$ – Jan Feb 13 '16 at 19:55

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